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%I #84 Dec 30 2023 23:46:01
%S 1,2,5,6,7,10,11,12,25,26,27,30,31,32,35,36,37,50,51,52,55,56,57,60,
%T 61,62,125,126,127,130,131,132,135,136,137,150,151,152,155,156,157,
%U 160,161,162,175,176,177,180,181,182,185,186,187
%N Positive numbers whose base-5 representation contains no 3 or 4.
%C 5 divides neither C(2s-1,s) = A001700(s) (nor C(2s,s) = A000984(s), central column of Pascal's Triangle) if and only if s is one of the terms in this sequence.
%C k such that binomial(2k,k) != 0 (mod 10). - _Benoit Cloitre_, Aug 18 2002
%C Let us recall the plan of Apery's irrationality proof. Consider the recurrence (n+1)^3 * u_(n+1) = (34n^3 + 51n^2 + 27n + 5)u_n - n^3 * u_(n-1). The solution with starting values u_0 = 1; u_1 = 5 has the peculiar property that it has integral terms, despite the fact that at every recursion step we divide by (n+1)^3. The n-th term is given by f(n) = Sum_{i=0..n} binomial(n+i,i)^2 * binomial(n,i)^2 = A005259(n) (see Beukers link) and m such that f(m) mod 5 <> 0 equals 2*a(m). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 08 2004
%C Numbers k such that A208279(k) <> 0. A073095 is a subsequence. - _Chai Wah Wu_, Dec 08 2023
%H Clark Kimberling, <a href="/A037453/b037453.txt">Table of n, a(n) for n = 1..1000</a>
%H Frits Beukers <a href="https://www.researchgate.net/publication/27708775_Consequences_of_Apery%27s_work_on_z3">Consequences of Apery's work on zeta(3)</a>, Rencontres Arithmétiques de Caen, zeta(3) irrationnel: les retombées, 1995.
%H Barry Brent, <a href="https://doi.org/10.20944/preprints202306.1164.v6">On the Constant Terms of Certain Laurent Series</a>, Preprints (2023) 2023061164.
%H W. Shur, <a href="https://doi.org/10.37236/1331">The last digit of C(2*n,n) and sigma C(n,i)*C(2*n-2*i,n-i)</a>, The Electronic Journal of Combinatorics, R16, Volume 4, Issue 2 (1997).
%F a(3n)=5a(n), a(3n+1)=5a(n)+1, a(3n+2)=5a(n)+2, where by definition a(0)=0. - _Emeric Deutsch_, Mar 23 2004
%F G.f. satisfies g(x) = 5*(1+x+x^2)*g(x^3) + (x + 2*x^2)/(1-x^3). - _Robert Israel_, Sep 02 2014
%e From _David A. Corneth_, Dec 23 2023: (Start)
%e 27_10 = 102_5 is a term since its base-5 representation contains no 3 and no 4.
%e 28_10 = 103_5 is not a term since its base-5 representation contains a 3.
%e (End)
%p a:= proc(t) option remember; 5*procname(floor(t/3))+ (t mod 3) end proc:
%p a(0):= 0:
%p seq(a(n),n=1..100); # _Robert Israel_, Sep 02 2014
%t Table[FromDigits[IntegerDigits[k,3],5], {k,60}] (* _T. D. Noe_, Apr 18 2007 *)
%t Rest[FromDigits[#,5]&/@Tuples[{0,1,2},4]] (* _Harvey P. Dale_, Aug 31 2016 *)
%t Select[Range[187], !Divisible[Binomial[2#, #], 10]&] (* _Stefano Spezia_, Dec 09 2023 *)
%o (PARI) f(n)=sum(i=0,n,binomial(n+i,i)^2*binomial(n,i)^2); for (i=1,1000,if(Mod(f(i),5)<>0,print1(i/2,",")))
%o (PARI) isok(k) = binomial(2*k, k) % 10; \\ _Michel Marcus_, Dec 08 2023
%o (PARI) is(n) = my(s = Set(digits(n, 5))); s[#s] < 3 \\ _David A. Corneth_, Dec 23 2023
%o (PARI) a(n) = fromdigits(digits(n, 3), 5) \\ _David A. Corneth_, Dec 23 2023
%o (Julia)
%o function a(n)
%o m, r, b = n, 0, 1
%o while m > 0
%o m, q = divrem(m, 3)
%o r += b * q
%o b *= 5
%o end
%o r end; [a(n) for n in 1:53] |> println # _Peter Luschny_, Jan 03 2021
%o (Python)
%o from itertools import count, islice
%o from sympy.ntheory.factor_ import digits
%o def A037453_gen(startvalue=1): # generator of terms >= startvalue
%o if startvalue <= 0: yield 0
%o yield from filter(lambda n: all(x<3 for x in digits(n, 5)[1:]), count(max(startvalue, 1)))
%o A037453_list = list(islice(A037453_gen(), 30)) # _Chai Wah Wu_, Dec 08 2023
%Y Cf. A050607, A005836, A007091.
%Y Cf. A000984, A073095, A208279.
%K nonn,easy,base
%O 1,2
%A _Clark Kimberling_
%E Better definition from _T. D. Noe_, Apr 18 2007
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